Question

我有一份清单清单：

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

我想在上面的列表中找到子列表的频率。

我试过使用itertools：

freq = [len(list(group)) for x in countall for key, group in groupby(x)]

然而，我得到了错误的结果：

[1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1]

我的列表理解有什么问题？

Answer 1

Groupby似乎处理彼此之后的序列。要使用它，您需要先对列表进行排序。另一种选择是使用Counter类：

from collections import Counter
countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]
Counter([tuple(x) for x in countall])

输出：

Counter({(3, 2): 10, (2, 3): 10, (1, 4): 5, (4, 1): 5, (5, 0): 1, (0, 5): 1})

Answer 2

如ForceBru所指出的那样首先对你的列表进行排序，然后使用groupby：

from itertools import groupby
countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4], [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

freq = [(key, len(list(x))) for key, x in groupby(sorted(countall))]
print(freq)

输出：

[([0, 5], 1), ([1, 4], 5), ([2, 3], 10), ([3, 2], 10), ([4, 1], 5), ([5, 0], 1)]

您的代码有错误：

freq = [len(list(group)) for x in countall for key, group in groupby(x)]
                       ^paranthesis missing

然后，您将countall中不需要的每个列表分组。

for x in countall for key, group in groupby(x)

你可以在排序（countall）

上直接groupby

另外，正如@Bemmu所回答，您可以使用collections.Counter。但是这不支持list所以首先你必须将数据转换为tupple或string然后使用Counter

Answer 3

如评论中所述，如果您使用的是groupby，则需要进行排序。

<强>代码：

import itertools as it
freq = {tuple(key): len(list(group)) for key, group in it.groupby(sorted(countall))}

测试代码：

countall = [[5, 0], [4, 1], [4, 1], [3, 2], [4, 1], [3, 2], [3, 2], [2, 3],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [4, 1], [3, 2], [3, 2], [2, 3], [3, 2], [2, 3], [2, 3], [1, 4],
           [3, 2], [2, 3], [2, 3], [1, 4], [2, 3], [1, 4], [1, 4], [0, 5]]

print(freq)

<强>结果：

{(3, 2): 10, (1, 4): 5, (2, 3): 10, (5, 0): 1, (0, 5): 1, (4, 1): 5}

计算列表中的频率

3 个答案: